Question

A bucket of height $8cm$ made up of a copper what is in the form of fraction of a right circular cone with the radii of its lower & upper are $3cm$ ands $9cm$ respectively find.
(i) the short height of the bucket
(ii) the volume of water and can be filled in bucket
(iii) the area of copper sheet required the make bucket.

Answer 1

We have,

$h=8cm$

$r_1=9cm$

$r_2=3cm$

Then, slant height

$L=\sqrt{8^2+6^2}=10cm$

$\left(i\right)$

Let $h_1$ be the height of the cone of which the bucket is a part of

$h_1=\frac{h{r}_1}{(r_1-r_2)}$

$=8\times \frac{9}{6}$

$=12cm$

$\left(ii\right)$

Volume of the water which can be filled in the bucket

$=\frac{8\pi (81+9+27)}{3}$

$=\frac{8\pi \times 117}{3}$

$=312\pi c{m}^3$

$\left(iii\right)$

Area of the metal sheet required to make the bucket $=$ surface area

$=\pi L(r_1+r_2)+\pi r_2^2$

$=\pi \times 10\times 12+9\pi$

$=129\pi c{m}^2$

Hence, this is the answer.